[QE-users] Magnetic anisotropy energy in QE 6.3

Thu Jul 12 13:57:56 CEST 2018

For the Co slab of the example I would suggest 20x20x1 kpoints for scf (no SOC) calculation and 40x40x1 kpoints for the nscf (with SOC) calculations.
But you should check convergence.

For bulk calculations MAE is extremely complicated to extract since in cubic systems the MAE is of the order of 10^(-5)eV!!! Almost impossible to achieve a converged result!

Cyrille

========================
Cyrille Barreteau
CEA Saclay, IRAMIS, SPEC Bat. 771
91191 Gif sur Yvette Cedex, FRANCE
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+33 1 69 08 38 56 /+33  6 47 53 66 52  (mobile)
email:     cyrille.barreteau at cea.fr
Web:     http://iramis.cea.fr/Pisp/cyrille.barreteau/
========================
________________________________
De : users [users-bounces at lists.quantum-espresso.org] de la part de Marcos Veríssimo Alves [marcos.verissimo.alves at gmail.com]
Envoyé : jeudi 12 juillet 2018 13:32
À : Quantum Espresso users Forum
Objet : Re: [QE-users] Magnetic anisotropy energy in QE 6.3

Thanks for the tip, Cyrille. Should it be increased in the SCF calculation or in the NSCF one? I would suppose the latter, but of course it's good to hear the opinion of an expert before wasting human and computer time.

Indeed, there's this paper by Daalderop et al. for bulk Fe, Ni and Co (if I'm not mistaken) where they highlight this - 100^3 k-points! I suppose that this would be equivalent to having a 100 x 100 x 100 Monkhorst-Pack grid?

>From your experience, maybe you could comment on the number of k-points needed for my particular calculations. The system am working on is a Fe adatom on a Cu2N surface over Cu(100). I am using a quite large vacuum layer, of about 18 Ang, along the z direction of a tetragonal cell - would a dense mesh of k-points then need to be considered only for the kx and ky directions of the BZ?

Best,

Marcos

On Thu, Jul 12, 2018 at 12:22 PM, BARRETEAU Cyrille <cyrille.barreteau at cea.fr<mailto:cyrille.barreteau at cea.fr>> wrote:
Of course to get reliable magnetic anisotropy you should drastically increase the number of Kpoints with respect to the example..

Cyrille

========================
Cyrille Barreteau
CEA Saclay, IRAMIS, SPEC Bat. 771
91191 Gif sur Yvette Cedex, FRANCE
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+33 1 69 08 38 56 /+33  6 47 53 66 52  (mobile)
email:     cyrille.barreteau at cea.fr<mailto:cyrille.barreteau at cea.fr>
Web:     http://iramis.cea.fr/Pisp/cyrille.barreteau/
========================
________________________________
De : users [users-bounces at lists.quantum-espresso.org<mailto:users-bounces at lists.quantum-espresso.org>] de la part de Marcos Veríssimo Alves [marcos.verissimo.alves at gmail.com<mailto:marcos.verissimo.alves at gmail.com>]
Envoyé : jeudi 12 juillet 2018 12:10

À : Quantum Espresso users Forum
Objet : Re: [QE-users] Magnetic anisotropy energy in QE 6.3

Hello Cyrille and Paolo,

Thanks for the quick response. I was interested really about the inner workings, which should be described in the paper that Cyrille mentioned. I will take a look at it. The actual execution of the calculation should be quite straightforward, from what I saw yesterday in the examples folder.

Once again, thanks to both of you for the response. If I run into any problems I'll ask, but hopefully all will be fine.

Best,

Marcos

Em qui, 12 de jul de 2018 10:37, BARRETEAU Cyrille <cyrille.barreteau at cea.fr<mailto:cyrille.barreteau at cea.fr>> escreveu:
Hi Marcos

The implementation of the Force Theorem has been described in the following paper:
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.90.205409

The procedure is the following:
first perform a scf calculation with scalar relativistic pseudo
then perform  nscf calculation with fully relativistic pseudo (option lforcet=.true., nosym=.true') starting from previous scf charge  (startingpot='file') with various spin orientations (theta=0,90 for example)
Finally perform a projwfc calculation with  lforcet=.true. and the value of the Fermi level from the nscf calculation (same ef_0 for all calculations).

Then you get a file with the energy decomposed over the various atoms and orbitals of the system..
The anisotropy is obtained by difference between the different spin orientations.

hope it helps..

Cyrille

========================
Cyrille Barreteau
CEA Saclay, IRAMIS, SPEC Bat. 771
91191 Gif sur Yvette Cedex, FRANCE
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+33 1 69 08 38 56 /+33  6 47 53 66 52  (mobile)
email:     cyrille.barreteau at cea.fr<mailto:cyrille.barreteau at cea.fr>
Web:     http://iramis.cea.fr/Pisp/cyrille.barreteau/
========================
________________________________
De : users [users-bounces at lists.quantum-espresso.org<mailto:users-bounces at lists.quantum-espresso.org>] de la part de Paolo Giannozzi [p.giannozzi at gmail.com<mailto:p.giannozzi at gmail.com>]
Envoyé : jeudi 12 juillet 2018 10:04
À : Quantum Espresso users Forum
Objet : Re: [QE-users] Magnetic anisotropy energy in QE 6.3

here? PP/examples/ForceTheorem_example/

P.

On Thu, Jul 12, 2018 at 10:01 AM, Marcos Veríssimo Alves <marcos.verissimo.alves at gmail.com<mailto:marcos.verissimo.alves at gmail.com>> wrote:
Hi all,

Browsing QE 6.3's documentation, I saw that MAE can be calculated as a post-processing step to a pw.x scf calculation. What is the exact procedure followed? I.e., in the post-processing calculation is the spin density rotated, and then SOC is included? Is there any reference that details the procedure used when the MFT is applied in Espresso?

Best regards,

Marcos

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